👨🏼‍🏫 🙋🏻 🔐 Cómo hacer un árbol de Navidad si eres matemático # 2 ⤵️ 🤝 📶

Continuación del artículo de ayer sobre fYolka a continuación.

Funciones básicas

Trapezoide

$y = \ left | x-4 \ right | + \ left | x + 2 \ right | -5.5$

Aquí el módulo del número se aplica dos veces, cambiando las constantes bajo el módulo y el valor sustraído, podemos ajustar la longitud del segmento con un valor constante y y el valor de y mismo en este segmento. Esta característica será útil más adelante para derrapes y cubos.

Elipse alternativa

$\ sqrt {\ left (x-1 \ right) ^ {2} +1,9 \ left (y-2 \ right) ^ {2}} - 1,3 = 0$

Notación de elipse alternativa. Las constantes dentro de los corchetes son responsables de las coordenadas del centro de la elipse, las constantes delante de los corchetes son para la relación de compresión a lo largo de los ejes, el número detrás de la raíz es el radio.

Elipse por dos puntos

$\ sqrt {\ left (x-1 \ right) ^ {2} + \ left (y-2 \ right) ^ {2}} + \ sqrt {\ left (x-0.1 \ right) ^ {2} + \ izquierda (y + 1 \ derecha) ^ {2}} - 3.3 = 0$

, - , (A B ) . .

, .

$\ max \ left (\ left | x \ right | -1, \ left | y \ right | -1 \ right) \ le0$

- :

$\ max \ left (\ left | x \ right |, \ left | y \ right | \ right) \ le1$

-

$s_ {1} = \ sqrt {\ left (x-10 \ right) ^ {2} +1.1 \ left (y-3.85 \ right) ^ {2}} - 0.55$ $s_ {2} = \ sqrt {\ left (x-10 \ right) ^ {2} +1.1 \ left (y-2.7 \ right) ^ {2}} - 0.85$ $s_ {3} = \ sqrt {\ left (x-10 \ right) ^ {2} +1.2 \ left (y-1.05 \ right) ^ {2}} - 1.15$

- min .

$\ min \ left (s_ {1}, \ s_ {2}, s_ {3} \ right) \ le0$

!

-

$- \ izquierda | x-1 \ derecha | - \ izquierda | x + 1 \ derecha | -y \ ge0$

- ,

$2-1.9 \ left | x-0.3 \ right | -1.9 \ left | x + 0.3 \ right | -y \ ge0$

, - 2 , - .

-

, .

$x=\frac{\left(\left|y\right|+y\right)}{2}$

$x=\frac{100\left(\left|y\right|-y\right)}{2}$

$x=\left(\frac{\left(\left|y\right|+y\right)}{2}+\frac{100\left(\left|y\right|-y\right)}{2}\right)$

$2-1.9\left|x-0.3\right|-1.9\left|x+0.3\right|-\left(\frac{\left(\left|y\right|+y\right)}{2}+\frac{100\left(\left|y\right|-y\right)}{2}\right)\ge0$

-

$2-1.9\left|x-9.7\right|-1.9\left|x-10.3\right|-\left(\frac{\left(\left|y-4\right|+y-4\right)}{2}+\frac{100\left(\left|y-4\right|-y+4\right)}{2}\right)\ge0$

$s_{4}=2-1.9\left|x-9.7\right|-1.9\left|x-10.3\right|-\left(\frac{\left(\left|y-4\right|+y-4\right)}{2}+\frac{100\left(\left|y-4\right|-y+4\right)}{2}\right)$ $\min\left(s_{1},\ s_{2},s_{3},-s_{4}\right)\le0$

s₄ , >0, <0, .

-

- x = 10, , , .

$h_{1}=\sqrt{\left(\left|x-10\right|\ -\ 0.8\right)^{2}+\left(y-2.7\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.8\right)^{2}+\left(y-2.5\right)^{2}}-2.015\\h_1\le0$

, = 10, = 2.55

$h_{2}=\sqrt{\left(\left|x-10\right|\ -\ 1.9\right)^{2}+\left(y-2.55\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.3\right)^{2}+\left(\left|y-2.55\right|-0.3\right)^{2}}-0.51\\h_2\le0$

$\min\left(s_{1},\ s_{2},s_{3},-s_{4},h_{1},h_{2}\right)\le0$

- 2

$100\left(\left|x-10\right|-0.2\right)^{2}+100\left(y-3.95\right)^{2}\le1$

-

$\left(300\left(\left|x-10\right|-0.03-0.-\left(y-3.6\right)\right)^{2}+3000\left(y-3.6\right)^{2}\right)\le1$

desmos

s_{1}=\sqrt{\left(x-10\right)^{2}+1.1\left(y-2.7\right)^{2}}-0.85

s_{2}=\sqrt{\left(x-10\right)^{2}+1.2\left(y-1.05\right)^{2}}-1.15

s_{3}=\sqrt{\left(x-10\right)^{2}+1.1\left(y-3.85\right)^{2}}-0.55

s_{4}=2-1.9\left|x-9.7\right|-1.9\left|x-10.3\right|-\left(\frac{\left(\left|y-4\right|+y-4\right)}{2}+\frac{100\left(\left|y-4\right|-y+4\right)}{2}\right)

h_{1}=\sqrt{\left(\left|x-10\right|\ -\ 0.8\right)^{2}+\left(y-2.7\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.8\right)^{2}+\left(y-2.5\right)^{2}}-2.015

h_{2}=\sqrt{\left(\left|x-10\right|\ -\ 1.9\right)^{2}+\left(y-2.55\right)^{2}}+\sqrt{\left(\left|x-10\right|\ -\ 2.3\right)^{2}+\left(\left|y-2.55\right|-0.3\right)^{2}}-0.51

\min\left(s_{1},\ s_{2},s_{3},-s_{4},h_{1},h_{2}\right)\le0

100\left(\left|x-10\right|-0.2\right)^{2}+100\left(y-3.95\right)^{2}\le1

\left(300\left(\left|x-10\right|-0.03-0.-\left(y-3.6\right)\right)^{2}+3000\left(y-3.6\right)^{2}\right)\le1

- . .

$d_{1}=-\left|x+7\right|-\left|x-14\right|+22\\d_{2}=\left|x+2.7\right|+\left|x-2.7\right|-6.35\\d_{3}=\left|x-9\right|+\left|x-11\right|-2.8$

$d=d_{1}+\left|d_{1}\right|+d_{2}-\left|d_{2}\right|+d_{3}-\left|d_{3}\right|$

$0.3d\left|\sin\left(13x\right)\right|$

d_{1}=-\left|x+7\right|-\left|x-14\right|+22

d_{2}=\left|x+2.7\right|+\left|x-2.7\right|-6.35

d_{3}=\left|x-9\right|+\left|x-11\right|-2.8

d=d_{1}+\left|d_{1}\right|+d_{2}-\left|d_{2}\right|+d_{3}-\left|d_{3}\right|

0.3d\left|\sin\left(13x\right)\right|

. - " ", , , x .

$\sqrt{\left|x\right|}+\sqrt{\left|y\right|}-0.45\le0$

mod,

$\left|\operatorname{mod}\left(x,2\right)-1\right|$

$f_{1}=\sqrt{\left|\operatorname{mod}\left(x,2\right)-1\right|}+\sqrt{\left|\operatorname{mod}\left(y,2\right)-1\right|}-0.45\\f_1\le0$

, .

$f_{2}=2xx+\left(y-6\right)^{2}-40\\f_{3}=2\left(x-10\right)^{2}+\left(y-2.5\right)^{2}-10\\f_2\le0\\f_3\le0$

$\min\left(-f_{1},f_{2},f_{3}\right)\ge0$

f_{1}=\sqrt{\left|\operatorname{mod}\left(x,2\right)-1\right|}+\sqrt{\left|\operatorname{mod}\left(y,2\right)-1\right|}-0.45

f_{2}=2xx+\left(y-6\right)^{2}-40

f_{3}=2\left(x-10\right)^{2}+\left(y-2.5\right)^{2}-10

\min\left(-f_{1},f_{2},f_{3}\right)\ge0

- . , |x| .

$j_{1}=\left|0.9\left|\left|x\right|-2.1\right|\right|-\left(y-1\right)-0.2\\j_1\le0$

$j_{2}=\left|\left|x\right|-2.1\right|^{2}-\left(y-1\right)^{2}-0.05\\j_2\ge0$

$j_{3}=0.2\left|\left|x\right|-2.1\right|^{2}+0.2\left(y-1\right)^{2}-0.1\\j_3\le0$

$j_{4}=\left(0.5\left|\left|x\right|-2.1\right|\right)^{2}+\left(y-1\right)^{2}-0.02\\j_4\le0$

:

$j_1j_4\le0$

$\max\left(j_{1}j_{4},\ -j_{2},\ j_{3}\right)\le0$

j_{1}=\left|0.9\left|\left|x\right|-2.1\right|\right|-\left(y-1\right)-0.2

j_{2}=\left|\left|x\right|-2.1\right|^{2}-\left(y-1\right)^{2}-0.05

j_{3}=0.2\left|\left|x\right|-2.1\right|^{2}+0.2\left(y-1\right)^{2}-0.1

j_{4}=\left(0.5\left|\left|x\right|-2.1\right|\right)^{2}+\left(y-1\right)^{2}-0.02

\max\left(j_{1}j_{4},\ -j_{2},\ j_{3}\right)\le0

, , .

$x_{1}=x\\y_{1}=y$

2021 MMXXI,

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t2,

$t_{2}=\max\left(\left|\left|x_{1}\right|-1\right|,\left|y_{1}-0.89\right|\right)-0.95\\t_{2}\le0$

"",

$\max\left(\left|1.2\left|x_{1}\right|-1.2\right|,\left|y_{1}-0.9\right|\right)-1\ge0$

V-

$\min\left(\left|2\left|x_{1}\right|-2\right|-y_{1},-\left|2\left|x_{1}\right|-2\right|+y_{1}+0.2,-t_{2}\right)\ge0$

$\max\left(\min\left(-t_{2},\max\left(\left|1.2\left|x_{1}\right|-1.2\right|,\left|y_{1}-0.9\right|\right)-1\right),\min\left(\left|2\left|x_{1}\right|-2\right|-y_{1},-\left|2\left|x_{1}\right|-2\right|+y_{1}+0.2,-t_{2}\right)\right)\ge0$

""

$\max\left(\left|\left|x_{1}\right|-1.05\right|,\left|y_{1}-0.9\right|\right)-1\le0$

$\min\left(\left|\left|x_{1}\right|-1.05\right|-\left|y_{1}-0.9\right|,\ -\left|\left|x_{1}\right|-1.05\right|+\left|y_{1}-0.9\right|+0.15\right)\ge0$

$\ max \ left (- \ min \ left (\ left | \ left | x_ {1} \ right | -1.05 \ right | - \ left | y_ {1} -0.9 \ right |, \ - \ left | \ left | x_ {1} \ right | -1.05 \ right | + \ left | y_ {1} -0.9 \ right | +0.15 \ right), \ max \ left (\ left | \ left | x_ {1} \ right | -1.05 \ right |, \ left | y_ {1} -0.9 \ right | \ right) -1 \ right) \ le0$

4.1 ,

$\ max \ left (- \ min \ left (\ left | \ left | x_ {1} -4.1 \ right | -1.05 \ right | - \ left | y_ {1} -0.9 \ right |, \ - \ left | \ left | x_ {1} -4.1 \ right | -1.05 \ right | + \ left | y_ {1} -0.9 \ right | +0.15 \ right), \ max \ left (\ left | \ left | x_ {1 } -4.1 \ derecha | -1.05 \ derecha |, \ izquierda | y_ {1} -0.9 \ derecha | \ derecha) -1 \ derecha) \ le0$

"I"

, "I"

$\ max \ left (\ left | x_ {1} -6.4 \ right | -0.06, \ left | y_ {1} -0.9 \ right | -01 \ right) \ le0$

t_{2}=\max\left(\left|\left|x_{1}\right|-1\right|,\left|y_{1}-0.89\right|\right)-0.95

\max\left(\min\left(-t_{2},\max\left(\left|1.2\left|x_{1}\right|-1.2\right|,\left|y_{1}-0.9\right|\right)-1\right),\min\left(\left|2\left|x_{1}\right|-2\right|-y_{1},-\left|2\left|x_{1}\right|-2\right|+y_{1}+0.2,-t_{2}\right)\right)\ge0

\max\left(-\min\left(\left|\left|x_{1}-4.1\right|-1.05\right|-\left|y_{1}-0.9\right|,\ -\left|\left|x_{1}-4.1\right|-1.05\right|+\left|y_{1}-0.9\right|+0.15\right),\max\left(\left|\left|x_{1}-4.1\right|-1.05\right|,\left|y_{1}-0.9\right|\right)-1\right)\le0

$x_ {1} = \ left (x \ -8 \ right) \ cdot1.3 \\ y_ {1} = \ left (y-9.3 \ right) \ cdot1.3$

, - , , , sin(x), x∈(-5, 5). .

$min = \ frac {f + g} {2} - \ left | \ frac {fg} {2} \ right | \\ max = \ frac {f + g} {2} + \ left | \ frac {fg} {2} \ right |$

Por lo tanto, el uso de las funciones mín. Y máx. En fórmulas de figuras es legal en esta tarea.

Cómo hacer un árbol de Navidad si eres matemático # 2

Funciones básicas

Trapezoide

Elipse alternativa

Elipse por dos puntos

-

!

-

- ,

-

-

-

- 2

-

desmos

:

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